Question

Figure shows three circuits, each consisting of a switch and two capacitors, initially charged as indicated. In which circuit the charge on the left hand capacitor will change after the closing of the switch ?

• A. 1
• B. 2
• C. 3
• D. All

Answer 1

Answer: (B), (C)

The correct options are
B 2
C 3

When two capacitors of charges $Q_1,Q_2$ of capacitances $C_1,C_2$ are connected.
Common potential $V_{com}=\frac{Q_1+Q_2}{C_1+C_2}$
Final charge on capacitor $C_1$ is $C_1\times V_{com}$

$\left(1\right)V_{com}=\frac{6q+3q}{2C+C}=\frac{3q}{c}$
$q_{2c}=\frac{3q}{C}\times 2C=6q$
The charge on the $2C$ capacitor remains the same.
$\left(2\right)V_{com}=\frac{6q+3q}{3C+C}=\frac{9q}{4C}$
$q_{3c}=\frac{9q}{4C}\times 3C=\frac{27q}{4}$
The charge on the capacitor $3C$ changes

$\left(3\right)V_{com}=\frac{6q+3q}{C+2C}=\frac{3q}{C}$
$q_c=\frac{3q}{C}\times C=3q$
The charge on the capacitor $C$ changes.

(OR)

Case(1)
Potential across $2C$ before closing switch $=\frac{6Q}{2C}=\frac{3Q}{C}$
Potential across $C$ before closing switch $=\frac{3Q}{C}$
As both capacitors have same potentials there will be no change in charges of capacitors after closing the switch.

Case(2)
Potential across $2C$ before closing switch $=\frac{6Q}{3C}=\frac{2Q}{C}$
Potential across $C$ before closing switch $=\frac{3Q}{C}$
As both capacitors have different potentials there will be change in charges of capacitors after closing the switch.

Case(3)
Potential across $2C$ before closing switch $=\frac{6Q}{C}$
Potential across $C$ before closing switch $=\frac{3Q}{2C}$
As both capacitors have different potentials, there will be change in charges of capacitors after closing the switch.